Trig Integral Practice Problem

Question

I solved this question using a different method than the answer book but the solutions don't produce the same result when we input x=1.

Answer 1

The calculus teacher's pet response: +C!

Recall that $\sin^2 x = 1 - \cos^2 x$...Can you find a similar formula for $\sin^4 x$?

For reference, here's what the two answers look like when graphed (dropping the +C term):

For another example of the same phenomenon, consider the integral $\int \sec^2 x \tan x d x$. Which of the two following answers is right?

1. using $u=\tan x$ and $du = \sec^2 x ~dx$, we have $$\int \sec^2 x \tan x ~dx = \int u ~du = \frac{1}{2} u^2 + C = \frac{1}{2} \tan^2 x +C$$

2. using $u=\sec x$ and $du = \sec x \tan x~ dx$ we have $$\int \sec^2 x \tan x dx = \int \sec x (\sec x \tan x)~ dx = \int u ~du = \frac{1}{2} u^2 +C = \frac{1}{2} \sec^2 x +C$$